Question

If the vectors \[2i - 3j\], \[i + j - k\]and \[3i - k\] form three concurrent edges of a parallelepiped, then the volume of the parallelepiped is
A) \[8\]
B) \[10\]
C) \[4\]
D) \[14\]

Answer 1

Hint: Parallelepiped is a \[3\] -D structure consisting of six symmetrical parallelograms. Here we are going to find the volume of parallelepiped which is found by using the scalar triple product concept. In this question we take the scalar triple product of vectors formed by three concurrent edges.

Formula used: Scalar triple product of vectors \[ = a.\left( {\vec b \times \vec c} \right)\]
Where and are three concurrent edges of parallelepiped.
\[\vec a{\rm{.}}\left( {\vec b \times \vec c} \right) = \left| a \right|\left| {\left( {\vec b \times \vec c} \right)} \right|\cos \left( \theta \right)\]
Where \[\theta \]is the angle between \[\vec a\]and \[\left( {\vec b \times \vec c} \right)\]
\[\vec a.\left( {\vec b \times \vec c} \right) = \left( {{a_1}\hat i + {a_2}\hat j + {a_3}\hat k} \right).\left| {\begin{array}{*{20}{c}}{\hat i}&{\hat j}&{\hat k}\\{{b_1}}&{{b_2}}&{{b_3}}\\{{c_1}}&{{c_2}}&{{c_3}}\end{array}} \right|\]

\[\vec a{\rm{.}}\left( {\vec b \times \vec c} \right) = \left( {{a_1}\hat i + {a_2}\hat j + {a_3}\hat k} \right)\]
\[ = \left( {{a_1}\hat i + {a_2}\hat j + {a_3}\hat k} \right).\left\{ {\left( {{b_2}{c_3} - {b_3}{c_2}} \right)\hat i - \left( {{b_1}{c_3} - {b_3}{c_1}} \right)\hat j + \left( {{b_1}{c_2} - {b_2}{c_1}} \right)\hat k} \right\}\]
\[ = {a_1}\left( {{b_2}{c_3} - {b_3}{c_2}} \right) - {a_2}\left( {{b_1}{c_3} - {b_3}{c_1}} \right) + {a_3}\left( {{b_1}{c_2} - {b_2}{c_1}} \right)\]

Complete step by step solution: Given: vectors a, b and c are the three concurrent edges of parallelepiped.
\[\vec a = 2\hat i - 3\hat j\]
\[\vec b = \hat i + \hat j - \hat k\]
\[\vec c = 3\hat i - \hat k\]
Where,
\[{a_1} = 2,{a_2} = - 3,{a_3} = 0\]
\[{b_1} = 1,{\rm{\;}}{b_2} = 1,{b_3} = - 1\]
\[{b_1} = 1,{\rm{\;}}{b_2} = 1,{b_3} = - 1\]


Volume of parallelepiped is given by :
\[\vec a.\left( {\vec b \times \vec c} \right) = \left( {{a_1}\hat i + {a_2}\hat j + {a_3}\hat k} \right).\left| {\begin{array}{*{20}{c}}{\hat i}&{\hat j}&{\hat k}\\{{b_1}}&{{b_2}}&{{b_3}}\\{{c_1}}&{{c_2}}&{{c_3}}\end{array}} \right|\]

\[ = \left( {{a_1}\hat i + {a_2}\hat j + {a_3}\hat k} \right).\left\{ {\left( {{b_2}{c_3} - {b_3}{c_2}} \right)\hat i - \left( {{b_1}{c_3} - {b_3}{c_1}} \right)\hat j + \left( {{b_1}{c_2} - {b_2}{c_1}} \right)\hat k} \right\}\]
\[ = {a_1}\left( {{b_2}{c_3} - {b_3}{c_2}} \right) - {a_2}\left( {{b_1}{c_3} - {b_3}{c_1}} \right) + {a_3}\left( {{b_1}{c_2} - {b_2}{c_1}} \right)\]
\[ = 2\left( { - 1} \right) + 3\left( 2 \right)\]
\[ = - 2 + 6\]
\[ = 4\]

Thus, Option (C) is correct.

Note: Here in this question we have to find the volume of parallelepiped. In order to find the volume of parallelepiped we must know the area of base and height of parallelepiped. Scalar triple product formula is used to find the volume of parallelepiped. Scalar triple product means product of three vectors i.e. dot product of one of the vectors with cross product of other two vectors.
Scalar triple products are represented as [a b c ].
The resultant scalar triple product is always scalar. Whenever we get the value of a scalar triple product as zero we can say that three vectors are coplanar.
Scalar triple product may be zero, negative and positive.